Question

The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is added to the first beaker and 100. mL of 8 M HCl is added to the second. (a) Find the ion-product constant for KCl at 20 °C. 14 Enter as a number to 2 decimal places. (b) What mass, if any, of KCl will precipitate from each beaker? Enter as a number to 0 decimal places. beaker 1: 0 grams beaker2: grams

Answer 1

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this: 

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

Answer 2

Answer:

(a) 13.69

(b) i beaker 1: 0g

    ii beaker 2: 0g

Explanation:

a. The solubility equilibrium equation for KCl is

KCl(s)  ⇄  K⁺(aq)  +  Cl⁻(aq)

3.7M KCl contains equal moles of K ions and Cl ions

therefore, the ion-product expression is written thus

Ksp = [K⁺][Cl⁻]

       = [3.7][3.7]

       = 13.69

b. from the first two beakers containing 100 mL and 3.7M KCl

moles of K⁺ = moles of Cl⁻ = moles of KCl = 3.7moles in 1L

if 3.7M Implies 3.7 moles in 1L or 1000 mL or 1000 cm³

how many moles will be contained in 100 mL

this is calculated as follows

3.7moles/Liter * 100 mL

$\frac{3.7 moles KCl}{1000 mL} * 100 ml = 0.37moles KCl$

= 0.37moles K⁺ = 0.37moles Cl⁻

4.0 M HCl, contains

$\frac{4 moles HCl}{1000 mL} *100mL = 0.4moles HCl = 0.4 moles H = 0.4moles Cl$ in 100mL

8.0M HCl, contains

$\frac{8moles HCL}{1000mL} *100mL=0.8mole HCl=0.8molesH=0.8molesCl$ in 100mL

now, in the first beaker 100 mL of 4M HCl is added to 100 mL of 3.7M KCl

total moles of Cl⁻ (0.4 + 0.37) moles = 0.77 moles

total moles of K⁺ remains 0.37 moles

total volume of solution = (100mL + 100mL) = 200mL/1000mL = 0.2L

total moles of Cl⁻ per Liter = 0.77moles/0.2L = 3.85M Cl⁻

total moles of K⁺ per Liter = 0.37moles/0.2L = 1.85M K⁺

Qsp must be greater or equal to Ksp for Precipitation to occur, that is

Qsp ≥ Ksp

Qsp = [K][Cl] = [1.85][3.85] = 7.12 this is less than 13.69(Ksp)

hence no KCl will precipitate in the first beaker

since there is no precipitate, there is therefore no need for calculating the mass precipitated

and the answer is 0g

(bii) now, in the second beaker 100 mL of 8M HCl is added to 100 mL of 3.7M KCl

total moles of Cl⁻ (0.8 + 0.37) moles = 1.17 moles

total moles of K⁺ remains 0.37 moles

total volume of solution = (100mL + 100mL) = 200mL/1000mL = 0.2L

total moles of Cl⁻ per Liter = 1.17moles/0.2L = 5.85M Cl⁻

total moles of K⁺ per Liter = 0.37moles/0.2L = 1.85M K⁺

Qsp must be greater or equal to Ksp for Precipitation to occur, that is

Qsp ≥ Ksp

Qsp = [K][Cl] = [1.85][5.85] = 10.82 this is less than 13.69(Ksp)

hence no KCl will precipitate also in the second beaker

since there is no precipitate, there is therefore no need fo calculating the mass precipitated

and the answer is 0g